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Andrei [34K]
4 years ago
12

Which biomes yearly rainfall mainly evaporates? i need a answer ASAP

Physics
1 answer:
makvit [3.9K]4 years ago
3 0
Desert. The desert doesn't get much rainfall and what it does get generally evaporates.
You might be interested in
Derive the following equations of motion
xz_007 [3.2K]

Answer:

___________________________________

<h3>a. Let us assume a body has initial velocity 'u' and it is subjected to a uniform acceleration 'a' so that the final velocity 'v' after a time interval 't'. Now, By the definition of acceleration, we have:</h3>

a =  \frac{v - u}{t}  \\ or \: at = v - u \\ v = u + at \:

It is first equation of motion.

___________________________________

<h3>b. Let us assume a body moving with an initial velocity 'u'. Let it's final body 'v' after a time interval 't' and the distance travelled by the body becomes 's' then we already have,</h3>

v = u + at...........(i) \\ s =  \frac{u + v}{2}  \times t.........(ii)

Putting the value of v from the equation (i) in equation (ii), we have,

s=  \frac{u + (u + at)}{2}  \times t \:  \: \\ or \: s =  \frac{(2u + at)t}{2}  \\ or \: s =  \frac{2ut + a {t}^{2} }{2}  \\ s = ut +  \frac{1}{2} a {t}^{2}

It is third equation of motion.

________________________________

<h3>c. Let us assume a body moving with an initial velocity 'u'. Let it's final velocity be 'v' after a time and the distance travelled by the body be 's'. We already have,</h3>

v = u + at.....(i) \\ s =  \frac{u + v}{2}  \times t......(ii) \\

v = u + at \\ or \: at = v - u \\ t =  \frac{v - u}{a}

Putting the value of t from (i) in the equation (ii)

s =  \frac{u + v}{2}  \times  \frac{v - u}{a}  \\ or \: s =   \frac{ {v}^{2}  -  {u}^{2} }{2a}  \\ or \: 2as =  {v}^{2}  -  {u}^{2}  \\  {v}^{2}  =  {u}^{2}  + 2as

It is forth equation of motion.

________________________________

Hope this helps...

Good luck on your assignment..

3 0
3 years ago
An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric
ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0
3 years ago
What are properties of a liquid?
Oksi-84 [34.3K]

Answer:

Pressure

Surface

Flow

Fluency

Explanation:

*Surface

The surface tension is the force with which the surface molecules of a liquid are attracted to bring them inside and thus decrease the surface area.

*Flow

It is defined as resistance to flow. The viscosity of a liquid depends on the intermolecular forces:

-The higher the intermolecular forces of a liquid, its molecules have a greater difficulty moving between them, therefore the substance is more viscous.

-Liquids that are made up of long, flexible molecules that can bend and tangle with each other are more viscous.

* Pressure

if the molecules of the liquid have a greater intensity of intermolecular force, then they will be trapped in the liquid and will have less facility to pass into the gas phase.

    on the contrary to lower intensity of intermolecular force, then the molecules can escape more easily to the gaseous state.

*Fluency

This property allows liquids to easily pass through a hole regardless of size, as long as this hole is at a lower or the same level of the container where the liquid is stored.

This property indicates the deformability of a liquid which is very wide without requiring mechanical stress.

6 0
4 years ago
Two uniform solid spheres, A and B have the same mass. The radius of sphere B is twice that of sphere A. The axis of rotation pa
Svetach [21]

Answer:

Sphere B has 4 times more inertia than sphere A.

Explanation:

Inertia on a Solid sphere is given by:

I = 2/5*M*R^2

For this problem:

ma = mb = M

Rb = 2*Ra

With these values:

Ia = 2/5*M*Ra^2

Ib = 2/5*M*(2*Ra)^2 = 4* (2/5*M*Ra^2)

As you can see, Ib = 4 * Ia.

4 0
3 years ago
A square coil, enclosing an area with sides 3.2 cm long, is wrapped with 2,720 turns of wire. A uniform magnetic field perpendic
marishachu [46]

Answer

\epsilon = -0.74\ V

Explanation:

given,                          

Side of square coil = 3.2 cm            

Number of turns = 2720                  

magnetic field in the coil increases to = 0.45 T

time interval = 1.7 s                  

average induced voltage in the coil = ?

length = 0.032 m                      

Area = 0.032² = 0.001024 m²

induced voltage                  

\epsilon = -NA \dfrac{dB}{dt}

\epsilon = -2720 \times 0.001024 \times \dfrac{0.45}{1.7}

\epsilon = -0.74\ V                          

The average voltage induced in the coil is equal to \epsilon = -0.74\ V                                              

3 0
3 years ago
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